bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

a)-7

b) 7

c) -2

d) 12

1) Ta có: \(x^2-2x-9y^2+6y\)

\(=x^2-2x+1-9y^2+6y-1\)

\(=\left(x-1\right)^2-\left(3y-1\right)^2\)

\(=\left(x-1-3y+1\right)\left(x-1+3y-1\right)\)

\(=\left(x-3y\right)\left(x+3y-2\right)\)

3) Ta có: \(x^2-9-4xy+4y^2\)

\(=\left(x-2y\right)^2-3^2\)

\(=\left(x-2y-3\right)\left(x-2y+3\right)\)

4) Ta có: \(\left(a+b\right)^2-\left(a-b\right)^2\)

\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)

\(=2b\cdot2a=4ab\)

5) Ta có: \(\left(x+y\right)^3-3xy\left(x+y\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)

6) Ta có: \(\left(x-y\right)^3+3xy\left(x-y\right)\)

\(=\left(x-y\right)\left[\left(x-y\right)^2+3xy\right]\)

\(=\left(x-y\right)\left(x^2-2xy+y^2+3xy\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\)

7) Ta có: \(4x^2-12x-46\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+9-55\)

\(=\left(2x-3\right)^2-55\)

\(=\left(2x-3-\sqrt{55}\right)\left(2x-3+\sqrt{55}\right)\)

Xin lỗi, mk chỉ biết bài 3:

Nhân cả 2 vế với 3 ta có:

3S = 1.2.3 +2.3.3 +3.4.3 +......+ 30.31.3

3S= 1.2.3 +2.3.( 4 - 1 ) +3.4. ( 5 - 2 ) +....+ 30.31. ( 32 - 29 )

3S= 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 +.....+ 30.31.32 - 30.31.29

3S= 30.31.32

S  = 30.31.32 : 3

S  = 9920

  Vậy S = 9920

cảm ơn bn nhé

có ai còn thức thì giúp mk làm bài 4 vs nhé.gấp lắm rồi

Bài 10 :

Câu a :

\(5xy\left(x-y\right)-2x+2y\)

\(=5xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(5xy-2\right)\)

Câu b :

\(6x-2y-x\left(y-3x\right)\)

\(=2\left(3x-y\right)+x\left(3x-y\right)\)

\(=\left(3x-2y\right)\left(2+x\right)\)

Câu c :

\(x^2+4x-xy-4y\)

\(=x\left(x+4\right)-y\left(x+4\right)\)

\(=\left(x+4\right)\left(x-y\right)\)

Câu d :

\(3xy+2z-6y-xz\)

\(=\left(3xy-6y\right)-\left(xz-2z\right)\)

\(=3y\left(x-2\right)-z\left(x-2\right)\)

\(=\left(x-2\right)\left(3y-z\right)\)

Bài 11 :

Câu a :

\(4-9x^2=0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy ........................

Câu b :

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy........................

Câu c :

\(2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..................

Câu d :

\(3x\left(x-4\right)-x+4=0\)

\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy................................

Câu e :

\(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy........................

Câu f :

\(\left(3x-y\right)^2-\left(x-y\right)^2=0\)

\(\Leftrightarrow\left(3x-y-x+y\right)\left(3x-y+x-y\right)=0\)

\(\Leftrightarrow2x\left(4x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy..........................

Câu 2: 

a: \(n^2-2n+5⋮n-1\)

\(\Leftrightarrow n^2-n-n+1+4⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)

b: \(4x^2-6x-16⋮x-3\)

\(\Leftrightarrow4x^2-12x+6x-18+2⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;2;5;1\right\}\)

Câu 3: 

a: \(\left(3x-8\right)\left(7x+10\right)-\left(2x-15\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(7x+10-2x+15\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(5x+25\right)=0\)

=>x=8/3 hoặc x=-5

b: \(\dfrac{\left(x^4-2x^2-8\right)}{x-2}=0\)(ĐKXĐ: x<>2)

\(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+2\right)=0\)

=>x+2=0

hay x=-2

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)